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2 years ago

The equation, DF + heat Imported Asset D + F, is an example of an endothermic reaction.

2 answers:

4 0

The answer is TRUE.

If the Energy is on the left, then the problem is true. If it is on the right then it would be negative, false, and considered as exothermic.

Endothermic reaction = the products are higher in energy than the reactants.
Exothermic reaction = a chemical reaction that releases energy by light or heat.

Readme [11.4K]2 years ago

4 0

Answer: It is true that the given example is an endothermic reaction.

Explanation:

Endothermic reaction is defined as the reaction in which reactant species absorb heat from the surrounding or from a source.

For example, $DF + Heat \rightarrow D + F$ is an endothermic reaction.

Energy of products is more than the energy of reactants in an endothermic reaction.

On the other hand, a reaction in which heat is released by reactant species is known as an exothermic reaction.

For example, $AB \rightarrow A + B + Heat$ is an exothermic reaction.

Energy of products is less than the energy of reactants in an exothermic reaction.

Hence, we can conclude that the given equation, $DF + Heat \rightarrow D + F$ , is an example of an endothermic reaction.

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Para que serve o balanceamento de equações químicas?

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2 years ago

Naproxen sodium, also known as Sodium 6-methoxy-α-methyl-2-napthaleneacetate, (C14H13NaO3), is an anti-inflammatory medication,

swat32

1) molar mass of C₁₄H₁₃NaO₃ = 252 g/mole

2) number of moles of naproxen sodium = 0.00119 moles

3) 0.00238 moles of naproxen in one day

Explanation:

1) Determine the molar mass of naproxen sodium.

We know that the molecular formula of naproxen sodium is C₁₄H₁₃NaO₃.

Now we can calculate the molar mass of naproxen sodium:

molar mass of C₁₄H₁₃NaO₃ = molar mass of C × 14 + molar mass of H × 13 + molar mass of Na × 1 + molar mass of O × 3

molar mass of C₁₄H₁₃NaO₃ = 12 × 14 + 1 × 13 + 23 × 1 + 16 × 3

molar mass of C₁₄H₁₃NaO₃ = 252 g/mole

2) Calculate the number of moles of naproxen sodium in a single tablet.

Knowing that one tablet contains 300 mg = 0.3 g of naproxen sodium, we use the following formula to find the number of moles:

number of moles = mass / molar weight

number of moles of naproxen sodium = 0.3 / 252

number of moles of naproxen sodium = 0.00119 moles

3) Calculate the number of moles of naproxen sodium that an adult would have taken if she took two doses of naproxen sodium in one day.

To find the number of moles taken by the adult in one day we follow the next reasoning:

if 1 tablet of naproxen contains 0.00119 moles of naproxen

then 2 tablets of naproxen contains X moles of naproxen

X = (2 × 0.00119) / 1 = 0.00238 moles of naproxen

Learn more about:

number of moles

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8 0

3 years ago

PLEASE HELP!!!!

stealth61 [152]

Hello
The answer is b
Have a nice day

7 0

3 years ago

Read 2 more answers

A solution is made by dissolving 0.0503 kg of biphenyl (C12H10) in

finlep [7]

Answer:

Letter A po

Explanation:

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5 0

2 years ago

"A gas turbine receives a mixture having the following molar analysis: 10% CO2, 19% H2O, 71% N2, at 720 K, 0.35 MPa and a volume

tino4ka555 [31]

Answer:

The answer is "2074.2 KW"

Explanation:

The mole part of $CO_2$ =0.1

The mole part of $H_2O$ = O.19

The mole part of $N_2$ =0.71

At temperature ( $T_1$ ) mixture receives from turbine =720K

At pressure ( $p_1$ ) mixture receives from turbine =0.35 Mpa

Flow rate volumetric V = 3.2 $\frac{m^3}{s}$

A turbine leaves the blend at temperature ( $T_2$ ) = 380K

The solution comes out of a pressure ( $p_2$ )= 0.11 Mpa

Decreasing healthy mass balance:

$W_{cv}= m(h_1-h_2)\\\\W_{cv}= m\frac{(\bar{h_1}-\bar{h_2})}{M_{mix}}\\\\W_{cv}= m\frac{Y_{co_2}(\bar{h_1}-\bar{h_2})_{co_2}+Y_{H_2o}(\bar{h_1}-\bar{h_2})_{H_2o}+Y_{N_2}(\bar{h_1}-\bar{h_2})_{N_2}}{M_{mix}}\\\\\ Mass \ flow \ rate:\\$

$m= \frac{(AV)_1}{V_1}\\\\m= \frac{(AV)_1 P_1}{(\bar{\frac{R}{M}})_{mix}V_1}\\\\\frac{m}{M_{mix}}= \frac{(AV_1)(P1)}{\bar{R}T_1}$

by increasing value we get:

$\frac{m}{M_{mix}}= \frac{8.2 \frac{kg}{s}(0.35 \times 10^6 \frac{N}{m^2})}{8.314 \frac{N.M}{kmol}(720 K)}$

After solve we get = 0.1871 $\frac{Kmol (mix)}{s}$

$W_{cv}= m\frac{Y_{co_2}(\bar{h_1}-\bar{h_2})_{co_2}+Y_{H_2o}(\bar{h_1}-\bar{h_2})_{H_2o}+Y_{N_2}(\bar{h_1}-\bar{h_2})_{N_2}}{M_{mix}}$

$T_1 =720K \ \ \bar h_1=28,211 \frac{KJ}{Kmol} \ \ co_2\\\\T_2 =380K \ \ \bar h_2=12,552 \frac{KJ}{Kmol} \ \ co_2\\\\T_1 =720K \ \ \bar h_1=24,840 \frac{KJ}{Kmol} \ \ H_2o\\\\T_2 =380K \ \ \bar h_2=12,672 \frac{KJ}{Kmol} \ \ H_2o\\\\$

$T_1 =720K \ \ \bar h_1=21,220 \frac{KJ}{Kmol} \ \ N_2\\\\T_2 =380K \ \ \bar h_2=11,055 \frac{KJ}{Kmol} \ \ N_2\\\\$

put the value in above given formula it will give $W_{cv}=$ 2074.2 KW

4 0

3 years ago